Angular Momentum and Angular Impulse

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AP Physics C: Mechanics › Angular Momentum and Angular Impulse

Questions 1 - 10
1

A spacecraft with $I=80\ \text{kg·m}^2$ must reach $\omega=0.60\ \text{rad/s}$ from rest; during a short burn, external torques are negligible except for the thrusters, so $J=\Delta L=I\Delta\omega$. Assume the torque direction matches the desired rotation. Considering the described system, calculate the angular impulse applied to the spacecraft during the interval.

48 N·m·s

48 N·m

0.48 N·m·s

133 N·m·s

Explanation

This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the spacecraft needs to change from rest (ωi = 0) to ωf = 0.60 rad/s, requiring a change in angular momentum. Choice A is correct because the angular impulse J = ΔL = IΔω = I(ωf - ωi) = 80 × (0.60 - 0) = 48 N·m·s. Choice B is incorrect because it might result from a calculation error or misunderstanding of the initial conditions, possibly assuming a different initial angular velocity. To help students: Emphasize careful reading of initial conditions (starting from rest means ωi = 0). Practice problems with various initial states to build pattern recognition. Watch for: assumptions about initial conditions and ensure students identify when objects start from rest versus already in motion.

2

An ice skater rotates about a vertical axis with $I_i=3.0\ \text{kg·m}^2$ and $\omega_i=2.0\ \text{rad/s}$. The skater pulls in arms to $I_f=1.2\ \text{kg·m}^2$ with negligible external torque, so $L=I\omega$ is conserved. Considering the described system, what is the final angular velocity of the system after the skater pulls in arms?

7.2 rad/s

5.0 rad/s

3.2 rad/s

0.80 rad/s

Explanation

This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is conserved in a closed system when no external torque acts, which is the key principle in this ice skater problem. In this problem, the ice skater changes their moment of inertia by pulling in their arms, with negligible external torque, so angular momentum remains constant. Choice B is correct because it accurately applies conservation of angular momentum: Li = Lf, so Iiωi = Ifωf, giving ωf = Iiωi/If = (3.0 × 2.0)/1.2 = 5.0 rad/s. Choice C is incorrect because it appears to use an incorrect calculation, possibly confusing the ratio of moments of inertia. To help students: Emphasize that angular momentum conservation is analogous to linear momentum conservation but involves I and ω. Encourage students to identify when external torque is negligible (like this skating problem). Watch for: confusion about when angular momentum is conserved versus when angular velocity is constant, and ensure understanding that decreasing I increases ω.

3

A wind turbine rotor with $I=2.0\times10^4\ \text{kg·m}^2$ spins at $\omega_i=0.80\ \text{rad/s}$. The wind drops and net torque becomes $\tau=-2.0\times10^3\ \text{N·m}$ for $t=4.0\ \text{s}$, using $\Delta L=\tau t$ and $L=I\omega$. Considering the described system, what is the final angular velocity of the system after 4.0 seconds?

8.0×10$^3$ kg·m$^2$/s

0.60 rad/s

1.0 rad/s

0.40 rad/s

Explanation

This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the wind turbine rotor experiences a negative net torque as the wind drops, which decreases its angular momentum and slows its rotation. Choice A is correct because it accurately calculates the final angular velocity: ΔL = τt = -2.0×10³ × 4.0 = -8.0×10³ kg·m²/s, then ωf = (Li + ΔL)/I = (2.0×10⁴ × 0.80 - 8.0×10³)/(2.0×10⁴) = (1.6×10⁴ - 8.0×10³)/(2.0×10⁴) = 0.60 rad/s. Choice B is incorrect because it appears to use an incorrect calculation, possibly making an error with the scientific notation. To help students: Emphasize careful handling of scientific notation in multi-step problems. Encourage students to verify that negative torque reduces angular velocity. Watch for: calculation errors with large numbers and scientific notation, and ensure understanding that wind turbines slow down when wind decreases.

4

A wind turbine rotor with $I=1.0\times10^3\ \text{kg·m}^2$ spins at $\omega_i=2.0\ \text{rad/s}$; the wind drops and the net torque becomes $\tau=-200\ \text{N·m}$ for $t=5.0\ \text{s}$, so $\Delta L=\tau t$ and $L=I\omega$. Neglect other torques. Based on the scenario above, what is the final angular velocity of the system after 5.0 seconds?

0.0 rad/s

1.0 rad/s

3.0 rad/s

2.2 rad/s

Explanation

This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the wind turbine experiences a negative torque of -200 N·m for 5.0 seconds, changing its angular momentum by ΔL = τt = -200 × 5.0 = -1000 N·m·s. Choice B is correct because the initial angular momentum is Li = Iωi = 1000 × 2.0 = 2000 N·m·s, the final angular momentum is Lf = Li + ΔL = 2000 + (-1000) = 1000 N·m·s, and the final angular velocity is ωf = Lf/I = 1000/1000 = 1.0 rad/s. Choice C is incorrect because it would mean the turbine stops completely, which would require exactly -2000 N·m·s of impulse, not -1000 N·m·s. To help students: Practice problems involving partial reduction of angular velocity versus complete stopping. Emphasize checking if the final result makes physical sense. Watch for: assumptions that negative torque always brings objects to rest, rather than just reducing angular velocity.

5

A solid disk with $I=0.80\ \text{kg·m}^2$ rotates at $\omega_i=12\ \text{rad/s}$. A motor applies constant torque $\tau=3.2\ \text{N·m}$ for $t=5.0\ \text{s}$, so $\Delta L=\tau t$. Based on the scenario above, what is the final angular velocity of the system after 5.0 seconds?

20 N·m

8.0 rad/s

20 rad/s

32 rad/s

Explanation

This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system, while angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the solid disk experiences a constant torque which changes its angular momentum, as described in the scenario. Choice C is correct because it accurately calculates the final angular velocity using the angular impulse-momentum theorem: ΔL = τt = 3.2 × 5.0 = 16 kg·m²/s, then ωf = (Li + ΔL)/I = (0.80 × 12 + 16)/0.80 = 20 rad/s. Choice B is incorrect because it appears to double the correct answer, possibly by misapplying the formula. To help students: Emphasize the step-by-step application of ΔL = τt and L = Iω. Encourage drawing before/after diagrams showing initial and final states. Watch for: confusion between torque and angular momentum units, and ensure students understand that angular impulse changes the angular momentum, not the angular velocity directly.

6

A rotating disk has $I=0.50\ \text{kg·m}^2$ and $\omega_i=16\ \text{rad/s}$. An external torque $\tau=1.5\ \text{N·m}$ acts for $t=4.0\ \text{s}$, so $\Delta L=\tau t$ and $L=I\omega$. Based on the scenario above, determine the change in angular momentum of the disk.

6.0 N·m

0.75 kg·m$^2$/s

32 kg·m$^2$/s

6.0 kg·m$^2$/s

Explanation

This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the rotating disk experiences an external torque which changes its angular momentum by the amount of the angular impulse applied. Choice B is correct because it accurately calculates the change in angular momentum using the angular impulse formula: ΔL = τt = 1.5 × 4.0 = 6.0 kg·m²/s. Choice C is incorrect because it gives the answer in N·m units (torque units) rather than kg·m²/s (angular momentum units), showing unit confusion. To help students: Emphasize that the change in angular momentum equals the angular impulse, regardless of initial conditions. Encourage dimensional analysis to catch unit errors. Watch for: confusion between torque and angular momentum units, and ensure students understand that ΔL is independent of the initial angular velocity for a given impulse.

7

A spacecraft has rotational inertia $I=120\ \text{kg·m}^2$ about its yaw axis and initially $\omega_i=0$; its thrusters provide a constant torque $\tau=30\ \text{N·m}$ for $t=4.0\ \text{s}$, with angular impulse $J=\tau t=\Delta L$. Ignore external torques from space. Based on the scenario above, calculate the angular impulse applied to the spacecraft during the interval.

120 N·m·s

7.5 N·m·s

120 N·m

480 N·m·s

Explanation

This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the spacecraft's thrusters provide a constant torque of 30 N·m for 4.0 seconds, creating an angular impulse. Choice B is correct because the angular impulse J = τt = 30 × 4.0 = 120 N·m·s, which equals the change in angular momentum ΔL. Choice C is incorrect because it gives only the torque value (30 N·m) multiplied by 4, confusing torque with impulse and missing the proper units. To help students: Emphasize that angular impulse has units of N·m·s (or kg·m²/s), not just N·m. Practice problems involving both impulse calculations and the resulting motion changes. Watch for: unit confusion between torque and angular impulse, and ensure understanding that impulse is the time integral of torque.

8

A bicycle wheel with moment of inertia $I=0.60\ \text{kg·m}^2$ starts from rest; a tangential force $F=18\ \text{N}$ is applied at the rim of radius $r=0.35\ \text{m}$ for $t=2.0\ \text{s}$, so $\tau=rF$ and $\Delta L=\tau t$. Neglect friction and assume the force stays perpendicular to the radius. Considering the described system, determine the change in angular momentum of the wheel.

25.2 kg·m$^2$/s

12.6 N

6.3 kg·m$^2$/s

12.6 kg·m$^2$/s

Explanation

This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the bicycle wheel experiences a tangential force at the rim, creating a torque τ = rF = 0.35 × 18 = 6.3 N·m for 2.0 seconds. Choice A is correct because the change in angular momentum is ΔL = τt = 6.3 × 2.0 = 12.6 kg·m²/s, using the impulse-momentum theorem for rotation. Choice B is incorrect because it only calculates the torque (6.3 N·m) without multiplying by time, a common error when students forget that impulse involves both force (or torque) and time duration. To help students: Reinforce the parallel between linear impulse (Ft) and angular impulse (τt). Encourage careful unit analysis to distinguish between torque (N·m) and angular momentum (kg·m²/s). Watch for: confusion between instantaneous quantities (torque) and time-integrated quantities (impulse).

9

A bicycle wheel with $I=0.40\ \text{kg·m}^2$ starts from rest; a constant tangential force $F=10\ \text{N}$ is applied at radius $r=0.30\ \text{m}$ for $t=5.0\ \text{s}$, so $\tau=rF$ and $\Delta L=\tau t$. Ignore friction and keep the force perpendicular to the radius. Based on the scenario above, determine the change in angular momentum of the wheel.

3.0 kg·m$^2$/s

6.0 kg·m$^2$/s

15 N·m

15 kg·m$^2$/s

Explanation

This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the bicycle wheel experiences a tangential force creating a torque τ = rF = 0.30 × 10 = 3.0 N·m for 5.0 seconds. Choice A is correct because the change in angular momentum is ΔL = τt = 3.0 × 5.0 = 15 kg·m²/s, properly applying the angular impulse-momentum theorem. Choice D is incorrect because it gives the answer in N·m instead of kg·m²/s, confusing the units of torque with those of angular momentum. To help students: Stress the importance of dimensional analysis in physics problems. Create unit conversion charts showing the relationships between rotational quantities. Watch for: unit confusion between torque (N·m) and angular momentum (kg·m²/s or N·m·s).

10

An ice skater spins with $I_i=3.0\ \text{kg·m}^2$ at $\omega_i=2.5\ \text{rad/s}$; they pull in their arms to $I_f=1.2\ \text{kg·m}^2$ while external torque is negligible, so $L=I\omega$ is conserved. Assume the change happens quickly compared with frictional effects. Considering the described system, what is the final angular velocity after pulling in their arms?

1.0 rad/s

6.25 rad/s

3.0 rad/s

2.0 rad/s

Explanation

This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the ice skater pulls in their arms with negligible external torque, so angular momentum is conserved: Li = Lf. Choice B is correct because using conservation of angular momentum: Iiωi = Ifωf, so 3.0 × 2.5 = 1.2 × ωf, giving ωf = 7.5/1.2 = 6.25 rad/s. Choice C is incorrect because it might result from incorrectly assuming the angular velocity changes by the same ratio as the moment of inertia, a common misconception about rotational dynamics. To help students: Emphasize that angular momentum (not angular velocity) is conserved when no external torque acts. Use analogies with linear momentum conservation in collisions. Watch for: confusion between conserved quantities and ensure students understand why angular velocity increases when moment of inertia decreases.

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